Square of a binomial

Multiply $(3x+5)$ by itself the long way and it takes four separate products before you can even start collecting terms. Do it a hundred times in an exam and the slow method costs you minutes you don't have. The square of a binomial identity turns that whole multiplication into a fixed three-term pattern you can write down on sight. This topic sits inside the Class 8 chapter Algebraic Identities, right after Algebraic Expressions, where the distributive law was used to multiply two binomials term by term. Here that same multiplication is applied to a binomial times itself, and the result is frozen into two identities worth memorising cold: $(a+b)^2$ and $(a-b)^2$ . Beyond expanding expressions, the identities double as a mental-maths shortcut for squaring numbers like $97$ or $102$ , and as a way to find $a^2+b^2$ from $a+b$ and $ab$ without ever solving for $a$ and $b$ . Multi-term squares like $(a+b+c)^2$ are Class 9 material and stay out of scope here.

The core insight

Squaring a binomial always gives three terms — square of the first term, plus or minus twice the product of the two terms, plus square of the second term — never just the sum of the two squares.

The core idea

A binomial has exactly two terms, such as $a+b$ or $3x-2y$ . Squaring it means multiplying the binomial by itself, and expanding that product with the distributive law gives a pattern that never changes.

Expanding $(a+b)^2$ . Write the square as a product and distribute:

(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2

where: $a$ and $b$ are the two terms of the binomial (numbers, variables, or both), $a^2$ and $b^2$ are their individual squares, and $2ab$ is twice their product — the middle term that is easy to forget.

Expanding $(a-b)^2$ . The same distribution with a subtraction gives:

(a-b)^2=(a-b)(a-b)=a^2-2ab+b^2

The only difference between the two identities is the sign of the middle term. The first and last terms, $a^2$ and $b^2$ , are always added and always positive, because a square can never be negative — even if $b$ itself is negative.

A connecting identity. Because both expansions share the same $a^2+b^2$ part, they can be combined to isolate it:

a^2+b^2=(a+b)^2-2ab=(a-b)^2+2ab

where: this lets you compute $a^2+b^2$ directly from $a+b$ (or $a-b$ ) and $ab$ , without ever finding the individual values of $a$ and $b$ . The cleanest way to see why the middle term is $2ab$ and not just $ab$ is a picture: a square of side $a+b$ splits into a smaller square of side $a$ , a smaller square of side $b$ , and two identical rectangles of sides $a$ and $b$ sandwiched between them.

Figure

A large square of side a plus b split into a corner square of area a squared, an opposite corner square of area b squared, and two equal rectangles of area a b each

Adding the four regions gives $a^2+ab+ab+b^2=a^2+2ab+b^2$ , the total area of the big square — the picture and the algebra agree term for term. This model is worth redrawing mentally whenever the identity feels abstract, and it also explains why the last term is always $b^2$ : it is literally the area of a square, which can never be negative. It also helps to see the two identities side by side, term by term, so the pattern is unmistakable:

Expression	First term squared	Middle term	Last term squared
$(a+b)^2$	$a^2$	$+2ab$	$b^2$
$(a-b)^2$	$a^2$	$-2ab$	$b^2$

The two rows differ in exactly one place — the sign in front of the middle term. Everything else, including the fact that both outer terms are squares and therefore always non-negative, is identical between the addition and subtraction cases. Keeping this table in mind is often faster than re-deriving the sign from scratch during an exam. These identities are not confined to pure algebra drills. In mensuration, if a square's side is given as a binomial such as $(x+4)$ cm, its area follows instantly as $(x+4)^2=x^2+8x+16$ square cm, with no measuring or redrawing required. Recognising the binomial-square pattern on sight — rather than reaching for full distributive multiplication every single time — is exactly the fluency this topic is meant to build, and it pays off again almost immediately: the very next chapter reverses these same identities to factorise perfect-square trinomials like $x^2+8x+16$ back into $(x+4)^2$ .

Derivation AProving the square-of-binomial identities by direct multiplication

Show step by step, starting only from the definition of squaring an expression and the distributive law learned for multiplying two binomials, that $(a+b)^2$ expands to exactly $a^2+2ab+b^2$ and that $(a-b)^2$ expands to exactly $a^2-2ab+b^2$ , so that both identities can be trusted as fixed patterns rather than re-derived every time they are used.

Step 1. Squaring a quantity means multiplying it by itself, so by definition $(a+b)^2=(a+b)\times(a+b)$ . This is nothing new — it is the same meaning of "squared" used for a single number, just applied to the two-term expression $a+b$ .

Step 2. Treat the first bracket as a single block and distribute it over the second bracket, the same way $(x+y)$ was multiplied by another binomial in the previous chapter: $(a+b)(a+b)=a(a+b)+b(a+b)$ .

Step 3. Distribute $a$ and $b$ separately into their brackets: $a(a+b)=a^2+ab$ and $b(a+b)=ba+b^2=ab+b^2$ . Adding these two results gives $a^2+ab+ab+b^2$ .

Step 4. Collect the two identical middle terms, $ab+ab=2ab$ , to reach the final form $a^2+2ab+b^2$ . This confirms $(a+b)^2=a^2+2ab+b^2$ with no assumption beyond ordinary multiplication.

Step 5. For subtraction, treat $a-b$ as $a+(-b)$ and repeat the same steps with $b$ replaced by $-b$ : $(a+(-b))^2=a^2+2a(-b)+(-b)^2=a^2-2ab+b^2$ , since $(-b)^2=b^2$ but $2a(-b)=-2ab$ . This proves $(a-b)^2=a^2-2ab+b^2$ , differing from the first identity only in the sign of the middle term.

Worked example

Expand $(3x+5)^2$ using the identity, verify by direct multiplication, then use the same identity to find $97^2$ mentally.

Step 1. Match $(3x+5)^2$ to the pattern $(a+b)^2=a^2+2ab+b^2$ with $a=3x$ and $b=5$ .

Step 2. Compute each piece: $a^2=(3x)^2=9x^2$ , the middle term $2ab=2(3x)(5)=30x$ , and $b^2=5^2=25$ .

Step 3. Add the pieces in order to get $(3x+5)^2=9x^2+30x+25$ .

Step 4. Verify by direct multiplication: $(3x+5)(3x+5)=3x(3x+5)+5(3x+5)=9x^2+15x+15x+25=9x^2+30x+25$ . Both methods agree, so $(3x+5)^2=9x^2+30x+25$ is confirmed.

Step 5. For $97^2$ , write $97=100-3$ and apply $(a-b)^2=a^2-2ab+b^2$ with $a=100$ , $b=3$ : $a^2=10000$ , $2ab=2(100)(3)=600$ , $b^2=9$ , giving $97^2=10000-600+9=9409$ . The identity turns a clumsy long multiplication into three easy squares and one subtraction, so $(3x+5)^2=9x^2+30x+25$ and $97^2=9409$ .

Competitive stretch

Given $a+b=12$ and $ab=32$ , find $a^2+b^2$ without solving for $a$ and $b$ . Use $a^2+b^2=(a+b)^2-2ab=12^2-2(32)=144-64=80$ . Checking against the actual roots $a=4,b=8$ : $16+64=80$ , matching exactly — the shortcut works even though $a$ and $b$ were never found individually.

Common traps

Writing $(a+b)^2=a^2+b^2$ and dropping the middle term entirely — always check the expansion has exactly three terms, not two.
Getting the sign of the middle term wrong in $(a-b)^2$ : it is $-2ab$ , but the last term $b^2$ stays positive, never $-b^2$ .
Forgetting to square the coefficient along with the variable: $(3x)^2=9x^2$ , not $3x^2$ .
Applying the identity to expressions with more than two terms, such as $(2x+3y-1)^2$ — that needs full distributive expansion, not this two-term identity, and is beyond Class 8 scope.
Trying to shortcut $2ab$ using rearrangements like $2ab=(a+b)^2-a^2-b^2$ without first mastering the basic expansion — always be able to fall back on direct multiplication to check any identity-based shortcut.

Ask the tutor: Want to see how these same identities get reversed to factorise perfect-square trinomials, or need more mental-squaring practice with numbers like $998^2$ ?

Why is the middle term always 2ab and not just ab?

Because $(a+b)^2$ is a product of two identical brackets, distribution produces the cross terms $ab$ twice — once as $a$ times $b$ from the first bracket, once as $b$ times $a$ from the second. Adding those two equal terms gives $2ab$ . It is not a rule to memorise separately; it falls straight out of multiplying $(a+b)(a+b)$ term by term, so re-deriving it once, as in the derivation above, is the best way to make it stick permanently rather than mixing it up with a plain $ab$ later.

Can I use this identity for something like 995 squared?

Yes, and that is exactly the kind of number it is built for. Write $995=1000-5$ , then apply $(a-b)^2=a^2-2ab+b^2$ with $a=1000$ and $b=5$ : $1000^2-2(1000)(5)+5^2=1000000-10000+25=990025$ . Choosing $a$ as a round number close to the target and $b$ as the small remaining difference is the standard trick for mental squaring, and the same trick works for numbers just above a round number too — for example $102^2=(100+2)^2=10000+400+4=10404$ .

Does this identity work if b is negative or a fraction?

Yes — $a$ and $b$ can be any real numbers, including negatives, fractions, or algebraic terms with variables. If $b$ is negative, substitute it with its sign; for example $(x-b)$ with $b<0$ actually becomes $(x+|b|)$ before squaring, so the middle term's sign flips accordingly once the actual value is plugged in. The identity itself never changes; only the arithmetic of substituting the actual values of $a$ and $b$ does, which is exactly why it is worth practising with a mix of whole numbers, negative numbers, and fractions before an exam.

Square of a binomial

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads